A boundary-preserving map on the unit disk We are given a (closed) ball $D^n$ and a (continuous) map $f: D^n \to D^n$, that is identity on the boundary of $D^n$. 
Let $C$ be a subset of $D^n$, and let $f^{-1}(C)$ be the inverse image of $C$ in $D^n$.
The claim is that there exists a map $g: C \to f^{-1}(C)$ that is identity on the intersection of $C$ and the boundary of $D^n$.
I actually suspect that the claim is wrong in general, but cannot find a counter-example. Also, if it is indeed wrong, what are the conditions on $C$ and $f$ so that it is correct? 
 A: This question is a real mess, due to all the posts by Petr before he figured out about comments. But if you read it through carefully, fedja has provided an answer, which I am copying here so this question will stop being bumped to the front page. The point is that the answer is no. We'll produce a set $C$ which will be a counterexample.
We start by defining a subset of $\mathbb{C}$. Let $A = [-1,0) \cup (0,1] \subset \mathbb{C}$. Fix a real number $0 < a < 1$, and consider the curve $x\mapsto x+ia\sin(\pi/x)$. Because $0 < a < 1$, we know this curve stays inside the unit disk. Define
$$B = \cup_{x \in A} (x+ia\sin(\pi/x)) \cup [−ia,ia]$$
Now take any continuous map $f$ that is identity on the boundary, sends $[−ia,ia]$ to one point (zero, for instance), and otherwise doesn't send any two points to one point (i.e. is injective away from $[-ia,ia]$). It is easy to construct such an $f$. Put $C=f(B)$. Then $C$ is a continuous path from $−1$ to $1$ but $B=f^{-1}(C)$ contains no such path, i.e. there is no continuous map $g$ from $[-1,1]$ to $B$ which is the identity at $-1$ and $1$. Therefore the claim is false.
